Show that $(x-\alpha)(x-\overline{\alpha})$ is a also a factor of $p(t)$ over the complex numbers [duplicate]

This question already has an answer here:

Here is the full question. Lots of struggles:

Let $p(t)$ belong to $P(R)$.

a) If $(x − \alpha)$ is a factor of $p(t)$ over the complex numbers (i.e. $p(t) = (x − \alpha)\cdot q(t)$, for $\alpha$ equal a complex number and for $q(t)$ a polynomial with coefficients in the complex numbers), then show that $(x − \alpha)(x - \overline{\alpha})$ is a also a factor of $p(t)$ over the complex numbers (where $\overline{\alpha}$ denotes the complex conjugate of $\alpha$)

and

b) Use the fundamental theorem of algebra to show that every polynomial with real number coefficients is the product of real polynomials of degree 1 or 2.

I've tried reading the book (Friedberg) and looking through my lecture notes to figure this out. I've tried using the unique factorization theorem for a) and, needless to say, have not had any luck (but this could also be because of the fresh "understanding" of the theorem). Anyone able to help? I just no idea how to go about proving this.

marked as duplicate by Nosrati, Namaste, Micah, Don Thousand, Adrian KeisterOct 5 '18 at 16:13

• Is there a missing assumption from a), that $p$ has real coefficients? Otherwise, the statement is false. – user61527 May 1 '14 at 0:48
• Yes I forgot to include this when typing up. Sorry about that and thank you for pointing it out. – user137087 May 1 '14 at 1:09

It’s not true as stated. First, the original polynomial $p$ has to have real coefficients. Second, the number $\alpha$ must be nonreal. You can easily construct counterexamples when either of these conditions is not satisfied.
But, granting these two additional hypotheses, it’s now not too hard. Consider the real polynomial $r(x)=(x-\alpha)(x-\overline\alpha)$. It’s a quadratic with no root in $\mathbb R$, and thus irreducible as a real polynomial. Now consider the real polynomials $r$ and $p$: I say that $r|p$. If not, then the two would be relatively prime, and you would have $Ap+Br=1$, an equation among real polynomials which is contradicted by plugging in $\alpha$ for the variable $x$. Thus $r$ and $p$ are not relatively prime, and so $r|p$.
For a) Show $p(\bar{t}) = \overline{p(t)}$ using the multiplicative nature of complex conjugates (here $\bar{a}$ denotes the complex conjugate of $a$).